Find last occurrence of period with regex - regex

I'm trying to create a regex for validating URLs. I know there are many advanced ones out there, but I want to create my own for learning purposes.
So far I have a regex that works quite well, however I want to improve the validation for the TLD part of the URI because I feel it's not quite there yet.
Here's my regex (or find it on regexr):
/^[(http(s)?):\/\/(www\.)?a-zA-Z0-9#:._\+~#=]{2,256}\.[a-zA-Z]{2,6}\b([/#?]{0,1}([A-Za-z0-9-._~:?#[\]#!$&''()*+,;=]|(%[A-Fa-f0-9]{2}))*)$/
It works well for links such as foo.com or http://foo.com or foo.co.uk
The problem appears when you introduce subdomains or second-level domains such as co.uk because the regex will accept foo.co.u or foo.co..
I did try using the following to select the substring after the last .:
/[(http(s)?):\/\/(www\.)?a-zA-Z0-9#:._\+~#=]{2,256}[^.]{2,}$/
but this prevents me from defining the path rules of the URI.
How can I ensure that the substring after the last . but before the first /, ? or # is at least 2 characters long?

From what I can see, you're almost there. Made some modification and it seems to work.
^(http(s)?:\/\/)?(www\.)?[a-zA-Z0-9#:._\+~#=]{2,256}\.[a-zA-Z]{2,6}([/#?;]([A-Za-z0-9-._~:?#[\]#!$&''()*+,;=]|(%[A-Fa-f0-9]{2}))*)?$
Can be somewhat shortened by doing
^(http(s)?:\/\/)?(www\.)?[\w#:.\+~#=]{2,256}\.[a-zA-Z]{2,6}([/#?;]([-\w.~:?#[\]#!$&''()*+,;=]|(%[A-Fa-f0-9]{2}))*)?$
(basically just tweaked your regex)
The main difference is that the parameter part is optional, but if it is there it has to start with one of /#?;. That part could probably be simplified as well.
Check it out here.
Edit:
After some experimenting I think this one is about as simple it'll get:
^(http(?:s)?:\/\/)?([-.~\w]+\.[a-zA-Z]{2,6})(:\d+)?(\/[-.~\w]*)?([#/#?;].*)?$
It also captures the separate parts - scheme, host, port, path and query/params.
Example here.

Related

Regex to not consider a phrase in a string

The background is as follows:
I m working on converting URL (with or without protocol and www) into
clickable links.
I have the regex working for URLs with http, https, ftp, file, www and some combination of http/https with www.
I also have the regex working for URLs with just www and no protocol.
However, I m unable to figure out a working one for finding URLs with no protocol and no server name (www).
I tried the following in (http://gskinner.com/RegExr/)
([^www\.|http\:// ][a-zA-Z0-9\.]+)((?:[a-zA-Z0-9]+\.)+)([a-zA-Z]{2,4})([\/a-zA-Z0-9]+)([\?][a-zA-Z0-9]+)?
But that seems to work only that website and not on my application. Any help is much appreciated.
OK, you're probably not going to like this answer much - but then maybe you will? I have a regular expression (adapted from ) that seems to find URLs in text. You can see a demo on regex101.com .
The actual expression is very very long - this is because it's got "every legal TLD (top level domain) in it, which is a good start for finding "good" URLs. Here it is
((?:(?:http|ftp|https):\/{2}){0,1}(?:(?:[0-9a-z_-]+\.)+(?:aero|asia|biz|cat|com|coop|edu|gov|info|int|jobs|mil|mobi|museum|name|net|org|pro|tel|travel|ac|ad|ae|af|ag|ai|al|am|an|ao|aq|ar|as|at|au|aw|ax|az|ba|bb|bd|be|bf|bg|bh|bi|bj|bm|bn|bo|br|bs|bt|bv|bw|by|bz|ca|cc|cd|cf|cg|ch|ci|ck|cl|cm|cn|co|cr|cu|cv|cx|cy|cz|cz|de|dj|dk|dm|do|dz|ec|ee|eg|er|es|et|eu|fi|fj|fk|fm|fo|fr|ga|gb|gd|ge|gf|gg|gh|gi|gl|gm|gn|gp|gq|gr|gs|gt|gu|gw|gy|hk|hm|hn|hr|ht|hu|id|ie|il|im|in|io|iq|ir|is|it|je|jm|jo|jp|ke|kg|kh|ki|km|kn|kp|kr|kw|ky|kz|la|lb|lc|li|lk|lr|ls|lt|lu|lv|ly|ma|mc|md|me|mg|mh|mk|ml|mn|mn|mo|mp|mr|ms|mt|mu|mv|mw|mx|my|mz|na|nc|ne|nf|ng|ni|nl|no|np|nr|nu|nz|nom|pa|pe|pf|pg|ph|pk|pl|pm|pn|pr|ps|pt|pw|py|qa|re|ra|rs|ru|rw|sa|sb|sc|sd|se|sg|sh|si|sj|sj|sk|sl|sm|sn|so|sr|st|su|sv|sy|sz|tc|td|tf|tg|th|tj|tk|tl|tm|tn|to|tp|tr|tt|tv|tw|tz|ua|ug|uk|us|uy|uz|va|vc|ve|vg|vi|vn|vu|wf|ws|ye|yt|yu|za|zm|zw|arpa)(?::[0-9]+)?(?:(?:\/(?:[~0-9a-zA-Z\#\+\%#\.\/_-]+))?(?:\?[0-9a-zA-Z\+\%#\/&\[\];=_-]+)?)?))
As you can see the vast majority of the expression is taken up making sure that the TLD is one of the many legal ones (270 alternatives. I didn't know there were this many until I stumbled on http://mathiasbynens.be/demo/url-regex where I found the seeds of this expression).
Changes I made to the expression I found at the link above - mostly I just made all the groups (except the outer one) non-capturing so there is just a single "match". In the sample I posted I showed that a "good" protocol definition (like http://) will be included in the capture, while a "bad" one (like http:/) will be ignored - however the following URL will still be captured. I also showed that adding punctuation right after the expression (tested with ; and !) doesn't phase the expression: it captures "up to that point" and not beyond.
Play with it and see how you like it. It is relatively poor (according to the above link) for "pathological" URLs, and doesn't work with Arabic etc - but I don't think, based on your question, that this would be an issue.
A short explanation:
(?:(?:http|ftp|https):\/{2}){0,1}
(?:http|ftp|https) - match one of http, ftp, or https - non capturing "OR" group
:\/{2} - followed by a colon and exactly two forward slashes
(?: …){0,1} - the whole thing zero or one times (so no protocol, or properly formed)
(?:(?:[0-9a-z_-]+\.)+
[0-9a-z_-]+\. - at least one of the characters in the given range, followed by a period
(?: )+ - the whole thing one or more times, non-capturing
(?:aero|asia …) - one of these strings, non-capturing (these are all the valid TLDs)
(?::[0-9]+)? - zero or one times a colon followed by one or more digits: port specification
- this makes sure that www.something.us:8080 is valid
Everything else that follows matches all the different things that can go after - directories, queries, etc.
#Floris - Your suggestion worked well. I edited it a little bit and utilized adding a # to detect emails as well. I also edited for a simpler workflow as well (without the TLD) -
((?:(?:http|ftp|https):\/{2}){0,1}(?:(?:[0-9a-z_#-]+\.)+(?:[0-9a-zA-Z]){2,4})(?::[0-9]+)?(?:(?:\/(?:[~0-9a-zA-Z\#\+\%\#\.\/_-]+))?(?:\?[0-9a-zA-Z\+\%#\/&\[\];=_-]+)?)?)
Thanks for the help.

RegEx filter links from a document

I am currently learning regex and I am trying to filter all links (eg: http://www.link.com/folder/file.html) from a document with notepad++. Actually I want to delete everything else so that in the end only the http links are listed.
So far I tried this : http\:\/\/www\.[a-zA-Z0-9\.\/\-]+
This gives me all links which is find, but how do I delete the remaining stuff so that in the end I have a neat list of all links?
If I try to replace it with nothing followed by \1, obviously the link will be deleted, but I want the exact opposite to have everything else deleted.
So it should be something like:
- find a string of numbers, letters and special signs until "http"
- delete what you found
- and keep searching for more numbers, letters ans special signs after "html"
- and delete that again
Any ideas? Thanks so much.
In Notepad++, in the Replace menu (CTRL+H) you can do the following:
Find: .*?(http\:\/\/www\.[a-zA-Z0-9\.\/\-]+)
Replace: $1\n
Options: check the Regular expression and the . matches newline
This will return you with a list of all your links. There are two issues though:
The regex you provided for matching URLs is far from being generic enough to match any URL. If it is working in your case, that's fine, else check this question.
It will leave the text after the last matched URL intact. You have to delete it manually.
The answer made previously by #psxls was a great help for me when I have wanted to perform a similar process.
However, this regex rule was written six years ago now: accordingly, I had to adjust / complete / update it in order it can properly work with the some recent links, because:
a lot of URL are now using HTTPS instead of HTTP protocol
many websites less use www as main subdomain
some links adds punctuation mark (which have to be preserved)
I finally reshuffle the search rule to .*?(https?\:\/\/[a-zA-Z0-9[:punct:]]+) and it worked correctly with the file I had.
Unfortunately, this seemingly simple task is going to be almost impossible to do in notepad++. The regex you would have to construct would be...horrible. It might not even be possible, but if it is, it's not worth it. I pretty much guarantee that.
However, all is not lost. There are other tools more suitable to this problem.
Really what you want is a tool that can search through an input file and print out a list of regex matches. The UNIX utility "grep" will do just that. Don't be scared off because it's a UNIX utility: you can get it for Windows:
http://gnuwin32.sourceforge.net/packages/grep.htm
The grep command line you'll want to use is this:
grep -o 'http:\/\/www.[a-zA-Z0-9./-]\+\?' <filename(s)>
(Where <filename(s)> are the name(s) of the files you want to search for URLs in.)
You might want to shake up your regex a little bit, too. The problems I see with that regex are that it doesn't handle URLs without the 'www' subdomain, and it won't handle secure links (which start with https). Maybe that's what you want, but if not, I would modify it thusly:
grep -o 'https\?:\/\/[a-zA-Z0-9./-]\+\?' <filename(s)>
Here are some things to note about these expressions:
Inside a character group, there's no need to quote metacharacters except for [ and (sometimes) -. I say sometimes because if you put the dash at the end, as I have above, it's no longer interpreted as a range operator.
The grep utility's syntax, annoyingly, is different than most regex implementations in that most of the metacharacters we're familiar with (?, +, etc.) must be escaped to be used, not the other way around. Which is why you see backslashes before the ? and + characters above.
Lastly, the repetition metacharacter in this expression (+) is greedy by default, which could cause problems. I made it lazy by appending a ? to it. The way you have your URL match formulated, it probably wouldn't have caused problems, but if you change your match to, say [^ ] instead of [a-zA-Z0-9./-], you would see URLs on the same line getting combined together.
I did this a different way.
Find everything up to the first/next (https or http) (then everything that comes next) up to (html or htm), then output just the '(https or http)(everything next) then (html or htm)' with a line feed/ carriage return after each.
So:
Find: .*?(https:|http:)(.*?)(html|htm)
Replace with: \1\2\3\r\n
Saves looking for all possible (incl non-generic) url matches.
You will need to manually remove any text after the last matched URL.
Can also be used to create url links:
Find: .*?(https:|http:)(.*?)(html|htm)
Replace: \1\2\3\r\n
or image links (jpg/jpeg/gif):
Find: .*?(https:|http:)(.*?)(jpeg|jpg|gif)
Replace: <img src="\1\2\3">\r\n
I know my answer won't be RegEx related, but here is another efficient way to get lines containing URLs.
This won't remove text around links like Toto mentioned in comments.
At least if there is nice pattern to all links, like https://.
CTRL+F => change tab to Mark
Insert https://
Tick Mark to bookmark.
Mark All.
Find => Bookmarks => Delete all lines without bookmark.
I hope someone who lands here in search of same problem will find my way more user-friendly.
You can still use RegEx to mark lines :)

Regex for URL routing - match alphanumeric and dashes except words in this list

I'm using CodeIgniter to write an app where a user will be allowed to register an account and is assigned a URL (URL slug) of their choosing (ex. domain.com/user-name). CodeIgniter has a URL routing feature that allows the utilization of regular expressions (link).
User's are only allowed to register URL's that contain alphanumeric characters, dashes (-), and under scores (_). This is the regex I'm using to verify the validity of the URL slug: ^[A-Za-z0-9][A-Za-z0-9_-]{2,254}$
I am using the url routing feature to route a few url's to features on my site (ex. /home -> /pages/index, /activity -> /user/activity) so those particular URL's obviously cannot be registered by a user.
I'm largely inexperienced with regular expressions but have attempted to write an expression that would match any URL slugs with alphanumerics/dash/underscore except if they are any of the following:
default_controller
404_override
home
activity
Here is the code I'm using to try to match the words with that specific criteria:
$route['(?!default_controller|404_override|home|activity)[A-Za-z0-9][A-Za-z0-9_-]{2,254}'] = 'view/slug/$1';
but it isn't routing properly. Can someone help? (side question: is it necessary to have ^ or $ in the regex when trying to match with URL's?)
Alright, let's pick this apart.
Ignore CodeIgniter's reserved routes.
The default_controller and 404_override portions of your route are unnecessary. Routes are compared to the requested URI to see if there's a match. It is highly unlikely that those two items will ever be in your URI, since they are special reserved routes for CodeIgniter. So let's forget about them.
$route['(?!home|activity)[A-Za-z0-9][A-Za-z0-9_-]{2,254}'] = 'view/slug/$1';
Capture everything!
With regular expressions, a group is created using parentheses (). This group can then be retrieved with a back reference - in our case, the $1, $2, etc. located in the second part of the route. You only had a group around the first set of items you were trying to exclude, so it would not properly capture the entire wild card. You found this out yourself already, and added a group around the entire item (good!).
$route['((?!home|activity)[A-Za-z0-9][A-Za-z0-9_-]{2,254})'] = 'view/slug/$1';
Look-ahead?!
On that subject, the first group around home|activity is not actually a traditional group, due to the use of ?! at the beginning. This is called a negative look-ahead, and it's a complicated regular expression feature. And it's being used incorrectly:
Negative lookahead is indispensable if you want to match something not followed by something else.
There's a LOT more I could go into with this, but basically we don't really want or need it in the first place, so I'll let you explore if you'd like.
In order to make your life easier, I'd suggest separating the home, activity, and other existing controllers in the routes. CodeIgniter will look through the list of routes from top to bottom, and once something matches, it stops checking. So if you specify your existing controllers before the wild card, they will match, and your wild card regular expression can be greatly simplified.
$route['home'] = 'pages';
$route['activity'] = 'user/activity';
$route['([A-Za-z0-9][A-Za-z0-9_-]{2,254})'] = 'view/slug/$1';
Remember to list your routes in order from most specific to least. Wild card matches are less specific than exact matches (like home and activity), so they should come after (below).
Now, that's all the complicated stuff. A little more FYI.
Remember that dashes - have a special meaning when in between [] brackets. You should escape them if you want to match a literal dash.
$route['([A-Za-z0-9][A-Za-z0-9_\-]{2,254})'] = 'view/slug/$1';
Note that your character repetition min/max {2,254} only applies to the second set of characters, so your user names must be 3 characters at minimum, and 255 at maximum. Just an FYI if you didn't realize that already.
I saw your own answer to this problem, and it's just ugly. Sorry. The ^ and $ symbols are used improperly throughout the lookahead (which still shouldn't be there in the first place). It may "work" for a few use cases that you're testing it with, but it will just give you problems and headaches in the future.
Hopefully now you know more about regular expressions and how they're matched in the routing process.
And to answer your question, no, you should not use ^ and $ at the beginning and end of your regex -- CodeIgniter will add that for you.
Use the 404, Luke...
At this point your routes are improved and should be functional. I will throw it out there, though, that you might want to consider using the controller/method defined as the 404_override to handle your wild cards. The main benefit of this is that you don't need ANY routes to direct a wild card, or to prevent your wild card from goofing up existing controllers. You only need:
$route['404_override'] = 'view/slug';
Then, your View::slug() method would check the URI, and see if it's a valid pattern, then check if it exists as a user (same as your slug method does now, no doubt). If it does, then you're good to go. If it doesn't, then you throw a 404 error.
It may not seem that graceful, but it works great. Give it a shot if it sounds better for you.
I'm not familiar with codeIgniter specifically, but most frameworks routing operate based on precedence. In other words, the default controller, 404, etc routes should be defined first. Then you can simplify your regex to only match the slugs.
Ok answering my own question
I've seem to come up with a different expression that works:
$route['(^(?!default_controller$|404_override$|home$|activity$)[A-Za-z0-9][A-Za-z0-9_-]{2,254}$)'] = 'view/slug/$1';
I added parenthesis around the whole expression (I think that's what CodeIgniter matches with $1 on the right) and added a start of line identifier: ^ and a bunch of end of line identifiers: $
Hope this helps someone who may run into this problem later.

Regex pattern to format url

I have this pattern ^(?:http://)?(?:www.)?(.*?)/?(.*?)$ but it's still not perfect.
Let's say we have these urls to test against it:
example.com
example.com/
www.example.com/
http://example.com/
example.com/param
http://example.com/params/
The final output should be example.com/ if there's no parameters and example.com/params/ if with parameters. My problem is that it matches only second group. It doesn't look like /? is working otherwise it would stop on slash character. Is it possible to achieve what I want using only one pattern?
So you want the host name in $1? Your regex is ambiguous, there are many ways to match it; the regex engine will prefer the longest, leftmost possible match. If you don't want slashes in the first part, then say so. Explicitly. (?:http://)?(?:www\.)?([^/]*)?/?(.*)?$
One that I've used is:
((?:(?:https?://)?[\w\d:##%/;$()~_?\+\-=&]+|www|ftp)\.[\w\d:##%/;$()~_?\+\-=&\.]+)
The problem with URLs is that there are SO many ways one can be written, which is why the above code looks so congested. This will match all your examples above, but it will also match things like:
alkasi.jaias
Hopefully this will get you headed to where you need or want to go, and perhaps someone might be able to come up behind me and clean it up some (it's early morning, I'm getting ready for work, and am exhausted. :P)

Writing Regular Expression for URL in Google Analytics

I have a huge list of URL's, in the format:
http://www.example.com/dest/uk/bath/
http://www.example.com/dest/aus/sydney/
http://www.example.com/dest/aus/
http://www.example.com/dest/uk/
http://www.example.com/dest/nor/
What RegEx could I use to get the last three URL's, but miss the first two, so that every URL without a city attached is given, but the ones with cities are denied?
Note: I am using Google Analytics, so I need to use RegEx's to monitor my URL's with their advanced feature. As of right now Google is rejecting each regular expression.
Generally, the best suggestion I can make for parsing URL's with a Regex is don't.
Your time is much much better spent finding a libary that exists for your language dedicated to the task of processing URLs.
It will have worked out all the edge cases, be fully RFC compliant, be bug free, secure, and have a great user interface so you can just suck out the bits you really want.
In your case, the suggested way to process it would be, using your URL library, extract the element s and then work explicitly on them.
That way, at most you'll have to deal with the path on its own, and not have to worry so much wether its
http://site.com/
https://site.com/
http://site.com:80/
http://www.site.com/
Unless you really want to.
For the "Path" you might even wish to use a splitter ( or a dedicated path parser ) to tokenise the path into elements first just to be sure.
tj111's current solution doesn't work - it matches all your urls.
Here's one that works (and I checked with your values). It also matches, no matter if there is a trailing slash or not:
http:\/\/.*dest\/\w+/?$
/http:\/\/www\.site\.com\/dest\/\w+\/?$/i
matches if they're all the same site with the "dest" there. you could also do this:
/\w+:\/\/[^/]+\/dest\/\w+\/?$/i
which will match any site with any protocal (http,ftp) and any site with the /dest/country at the end, and an optional /
Note, that this will only work with a subset of what the urls could legitimately be.
Try this regular expression:
^http://www\.example\.com/dest/[^/]+/$
This would only match the last three URLs.